Gottfried Wrote:Well in this case, for the result s being real only, alpha is a unique function of beta. If you take +beta you need one alpha, and if you take -beta, you need another alpha adapted from this; in this case, the two alphas are even equal (the graph looks symmetrical w.r.t. y-axis).
So no ambiguity is here, just use the formula of my paper, by which the alpha is determined by beta to get a real s.

Gottfried

I do not take +- beta, I take +- I . beta is still 0<beta<pi. In that case, IMAG(t) at beta=pi/2 would be -I, so graph will be on the right lower quadrant. However, at beta - pi/2, if we take +i, we get IMAG(t) = -i -> in the left lower quadrant, while at -I it will be where it is now-left upper quadrant.

The question is, can we even use beta = - pi/2 as that would imply h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i.

Gottfried Wrote:Well in this case, for the result s being real only, alpha is a unique function of beta. If you take +beta you need one alpha, and if you take -beta, you need another alpha adapted from this; in this case, the two alphas are even equal (the graph looks symmetrical w.r.t. y-axis).
So no ambiguity is here, just use the formula of my paper, by which the alpha is determined by beta to get a real s.

Gottfried

I do not take +- beta, I take +- I . beta is still 0<beta<pi.

But that is funny. What else is u = alpha + beta*I , so to have some U with imaginary part 1*I or -1*I, you use beta = 1 or beta = -1 .
And then see, what value the real part must have such that the resulting s is purely real.
And while the graph shows only u = alpha -pi*I .. u=alpha + pi*I there is no problem to extend the graph to u = alpha + (beta + 2*k*pi)*I - there is a rough periodicity with a distortion then. Just plot it in other ranges.

Quote:In that case, IMAG(t) at beta=pi/2 would be -I, so graph will be on the right lower quadrant. However, at beta - pi/2, if we take +i, we get IMAG(t) = -i -> in the left lower quadrant, while at -I it will be where it is now-left upper quadrant.

The question is, can we even use beta = - pi/2 as that would imply

Yes, just look in the 3'rd quadrant; the required value for alpha is on the blue curve, and the resulting values for real and imag of t are on the magenta curves, and the resulting value for s is on the green curve.

Quote:h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i.

Gottfried Wrote:But that is funny. What else is u = alpha + beta*I , so to have some U with imaginary part 1*I or -1*I, you use beta = 1 or beta = -1 .
And then see, what value the real part must have such that the resulting s is purely real.

What I mean is that any time You define an arbitrary complex number u you have to consider both roots of (-1) and only later get rid of one if that does not add any information. Usually it does not, so people already initially dismiss the idea of 2 roots of sqrt(-1). Euler did not, however, and that is why he never made mistakes.

Which means that beta is 1, but sqrt(-1) is -i and +i simultaneously not because of beta, but because of 2 values of sqrt(-1) which can occur simultaneously for any given beta.

If You take beta = - 1 , You get u=a-sgrt(-1) but sqrt (-1) again taken as usual is +-i, so You have a-+i instead of a+- i as in case of b=1. Not that it matters, but it matters in your graphs, as by using +-i instead of +- beta You will get 2 values of IMAG(t) = +i and -i in the right side of Y axis, meaning 2 symmetric curves there.

Quote:And while the graph shows only u = alpha -pi*I .. u=alpha + pi*I there is no problem to extend the graph to u = alpha + (beta + 2*k*pi)*I - there is a rough periodicity with a distortion then. Just plot it in other ranges.

Could You please plot few examples on the current plot just to get the idea-it is very interesting to see the character of those branches- I am really sorry, I have only excel as a tool and even that I use for business tables not for graphing functions. So please...

Quote:h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i.

I think, may be wrongly, that beta in Your case corresponds to the power of e in tetration of (e^beta). If not, please correct me.

So I think if beta = pi/2 it is Ok, as h(e^pi/2) diverges and has 2 values +I , -I and ONLY then other branches, so e^pi/2 is a real value of s that corresponds to complex to real transformation as s>e^1/e. You should be able to see both I and - I at beta = pi/2.

But if beta = -pi/2 something is wrong as (e^-pi/2) = 0,20..and
(e^-pi/2)^(e^-pi/2)^(e^-pi/2) ....... converges, and is 0,47..... which means that is real to real case where beta should have been 0, not -pi/2- there should not have been such branch on a graph in this quadrant.

So for me it looks that if You just fold the graph along Y so that left side superimposes on right side You will have the right graph+ have left side free to plot convergent values of h(e^beta) where e^beta< e^1/e so beta < 1/e (including negative beta).

May be I misunderstand something still in the role of each variable, but, if You will be able to explain it to me, others will easily understand

Gottfried Wrote:But that is funny. What else is u = alpha + beta*I , so to have some U with imaginary part 1*I or -1*I, you use beta = 1 or beta = -1 .
And then see, what value the real part must have such that the resulting s is purely real.

What I mean is that any time You define an arbitrary complex number u you have to consider both roots of (-1) and only later get rid of one if that does not add any information. Usually it does not, so people already initially dismiss the idea of 2 roots of sqrt(-1). Euler did not, however, and that is why he never made mistakes.

Which means that beta is 1, but sqrt(-1) is -i and +i simultaneously not because of beta, but because of 2 values of sqrt(-1) which can occur simultaneously for any given beta.

If You take beta = - 1 , You get u=a-sgrt(-1) but sqrt (-1) again taken as usual is +-i, so You have a-+i instead of a+- i as in case of b=1. Not that it matters, but it matters in your graphs, as by using +-i instead of +- beta You will get 2 values of IMAG(t) = +i and -i in the right side of Y axis, meaning 2 symmetric curves there.

But, wait a moment. If you want I and -I in your formula for any variable x (or in my case u), what else are you doing as considering
sqrt(-1)_1 = 0 + b*I
and sqrt(-1)_2 = 0 - b*I
when you ask for considering both complex roots of -1 ?
So in my negative and positive beta I'm just doing that, what you demand; it is the negative and positive b in the above example.

So since it seems to me, that I'm already doing what you ask for, this is the source of my not-understanding of your concern.

Quote:I think, may be wrongly, that beta in Your case corresponds to the power of e in tetration of (e^beta). If not, please correct me.

How comes that you think this? I say: h(x) has many solutions t_0, t_1, t_2, ... So what is the form of the different t? Then I say, t_0,t_1,t_2 must be an exponential of u_0,u_1,u_2,... . complex, multivalued like your example of sqrt(-1). Now: what is the complex form of all these t's? They must be the exponential of another number u_0,u_1,u_2,... multivalued again. Now what is the form of the u_0,u_1,u_2,..., which are complex numbers alpha_0 + beta_0*I, alpha_1 +beta_1*I, maybe each second beta is just the negative signed other beta, which would satisfy your request to consider positive and negative complex roots.

Then I find, that if I select one imaginary part beta*I, to get a real s, the real part alpha must be of a certain value. I can insert beta_0*I, -beta_0*I and so on, compute the required alpha and always get one real s.
Your demand for considering -1*I and +1*I as possible roots of -1 is perfectly modeled here - at least as far I can see.

Quote:But if beta = -pi/2 something is wrong as (e^-pi/2) = 0,20..and
(e^-pi/2)^(e^-pi/2)^(e^-pi/2) ....... converges, and is 0,47..... which means that is real to real case where beta should have been 0, not -pi/2- there should not have been such branch on a graph in this quadrant.

My process goes three steps:
beta*I -> u -> exp(u) -> exp(u/exp(u)) = s = real(s)
and beta is not in that way related to s as you assume here. (It is not s = e^beta, for instance)

Gottfried Wrote:But, wait a moment. If you want I and -I in your formula for any variable x (or in my case u), what else are you doing as considering
sqrt(-1)_1 = 0 + b*I
and sqrt(-1)_2 = 0 - b*I
when you ask for considering both complex roots of -1 ?
So in my negative and positive beta I'm just doing that, what you demand; it is the negative and positive b in the above example.
So since it seems to me, that I'm already doing what you ask for, this is the source of my not-understanding of your concern.

Now I understand- and that is fine- I was just distracted by that (-pi/2) on X axis giving the same value for s=e^pi/2=4,81 (graphically it looks so ) . So that -pi/2 is just the same b=pi/2 multiplied by -1 which could be looked upon as taking the other root of (-1). OK!

Quote:Your demand for considering -1*I and +1*I as possible roots of -1 is perfectly modeled here - at least as far I can see.

Good! Now I think I will be finally able to place myself in the graph...
On other hand, then it is just a coincidence that having b= pi/2 the result for IMAG(t) = +-i coincides with h(e^pi/2) = +- i .

But please could You show a plot with few more branches of IMAG(t) on it? I am very interested what patterns You described look like now when I seem to understand where on the plot what is placed

[quote=Gottfried]
Here is a more extended plot showing a wider range for beta.
The periodicity of alpha and s is not perfect; the same real s approximate the vertical axes at multiples of pi.

I omitted the curves for t (= a + b*i) here.
[\quote]

Thanks. Very interesting. But can You add t=a+b*i as well? It would be easier for me to place myself on this graph, when I see the exact relation with the smaller graph I hope I have understood.

Why log(log(s)) = alpha =0 at beta= pi/2+-n*pi? Is it exact relation? Would that mean that s= e there? what is IMAG( t) in these points? They seem to be roots of d^2 (log(log(s))/d(beta)^2=0 and d^2(alpha) /dbeta^2 = 0(except for n=0?) .
I rememmber on the small graph, Imag (t) had inflection points at b=pi/2 while alpha=real (t) had inflection points are integer beta=1. Does it continue, the Imag (t) and real(t) to oscillate around x axis with decreasing amplitude as beta gets bigger? How does their max amplitude decay with beta- what is the functional dependance?

Is there any information in a curve connecting points with the same slope in this graph on different branches? Where

d/(dbeta ) of (log(log(s) ) = const? ( and in d(alpha)/dbeta=const)

what is the value of alpha max in interval where beta<pi/2, so that alpha(beta=0) = ? And what is s at beta=0 ? a at beta =0 was 3, right?

Ivars Wrote:My feeling is that the whole number axis except transcendental numbers like e,pi, 2 can be constructed from infinite tetration-may be that is how nature works?

I don't know whether this fits your question, but look at the postings of Gianfranco [GFR]. He stated, that every (real? complex?) number can be expressed as a powertower to a base b ( and a topexponent x ?) - don't have the link to this statement at hand.